Question: For Problem 10.17, show (as in Problem 1) that the Green function is

$\mathit{G}\left(t,t^{\text{'}}\right)\mathbf{=}\{\begin{array}{ll}0,& 0<t<t^{\text{'}}\\ (1/a)sinha\left(t-t^{\text{'}}\right),& 0<t^{\text{'}}<t\end{array}$

Thus write the solution of Problem 10.17as an integral [similar to (12.6)] and evaluate it.

The given expressions are $y^{\text{'}\text{'}}-a^{2}y=f\left(t\right)\delta \left(t^{\text{'}}-t\right)$.

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandanti-derivative.

In order to find the green function to the differential equation given by

$y^{\text{'}\text{'}}-a^{2}y=f\left(t\right)$

Where, we are giving that the boundary values for such differential equation, is

$$\begin{gathered} y_0^{\text{'}}=y_0=0 \\ f\left(t\right)={\int }_0^{\infty }f\left(t^{\text{'}}\right)\delta \left(t^{\text{'}}-t\right)d{t}^{\text{'}} \end{gathered}$$

Then, the solution to the given differential equation, is given in terms of the green function, as follows

$y\left(t\right)={\int }_0^{t}G\left(t,t^{\text{'}}\right)f\left(t^{\text{'}}\right)d{t}^{\text{'}}$

And, the green function is obtained by solving the following equation,

$\frac{d^2}{d{t}^2}G\left(t,t^{\text{'}}\right)-a^{2}G\left(t,t^{\text{'}}\right)=\delta \left(t^{\text{'}}-t\right)$

Where, by taking the Laplace inverse of this equation, we get

$L\left\{\frac{d^2}{d{t}^2}G\left(t,t^{\text{'}}\right)\right\}=p^{2}L\left\{G\left(t,t^{\text{'}}\right)\right\}-pG\left(0,t^{\text{'}}\right)-\frac{d}{dt}G\left(0,t^{\text{'}}\right)$

Where, we know that

$\delta \left(t^{\text{'}}-\tau \right)=\delta \left(\tau -t^{\text{'}}\right)$

then the value of the integral is

$G\left(t,t^{\text{'}}\right)=\frac{1}{a}\sinh \left(a\left(t-t^{\text{'}}\right)\right)$

Else, the integral is zero, thus we have

$G\left(t,t^{\text{'}}\right)=\left\{\begin{array}{ll}\frac{1}{a}\sinh \left(a\left(t-t^{\text{'}}\right)\right),& t>t^{\text{'}}\\ 0,& t<t^{\text{'}}\end{array}\right.$

Knowing, the green function, we can now find the solution to the differential equation given, where we know that the function f(t') is given by

$f\left(t^{\text{'}}\right)=\left\{\begin{array}{ll}0,& t^{\text{'}}<0\\ 1,& t^{\text{'}}>0\end{array}\right.$

Thus, the solution to the given differential equation in the integral form, is thus

$y\left(t\right)={\int }_0^t\frac{1}{a}\sinh \left(a\left(t-t^{\text{'}}\right)\right)f\left(t^{\text{'}}\right)d{t}^{\text{'}}$

Hence, evaluating the integral in case, we get

$\begin{array}{c}y\left(t\right)={\int }_0^t\frac{1}{a}\sinh \left(a\left(t-t^{\text{'}}\right)\right)d{t}^{\text{'}}\\ =\frac{1}{a}\left({\left.\frac{\cosh \left(a\left(t-t^{\text{'}}\right)\right)}{-a}\right|}_t^0\right.\\ =\frac{1}{a}\left(\frac{\cosh (a\times 0)}{-a}+\frac{\cosh \left(a\right(t-0\left)\right)}{a}\right)\\ =\frac{1}{a}\left(\frac{-1}{a}+\frac{\cosh \left(at\right)}{a}\right)=\frac{\cosh \left(at\right)-1}{a^2}\end{array}$

Thus, the solution to the given differential equation is

$y\left(t\right)=\left\{\begin{array}{ll}\frac{\cosh \left(at\right)-1}{a^2},& t>0\\ 0,& t<0\end{array}\right.$